**Exercise 2** 1. **Show that 𝓡 is an order relation on ℝ³.** An order relation must be reflexive, antisymmetric, and transitive. - Reflexive: For any $(x,y,z) \in \mathbb{R}^3$, check if $(x,y,z)𝓡(x,y,z)$. $$|x-x| = 0 \le y - y = 0 \quad \text{and} \quad z = z$$ So $(x,y,z)𝓡(x,y,z)$ is true. Reflexivity holds. - Antisymmetric: Suppose $(x,y,z)𝓡(a,b,c)$ and $(a,b,c)𝓡(x,y,z)$. Then $$|x - a| \le b - y \quad \text{and} \quad z = c$$ $$|a - x| \le y - b \quad \text{and} \quad c = z$$ Since $z=c$ and $c=z$, these hold. From inequalities we get: $$|x - a| \le b - y \quad \text{and} \quad |x - a| \le y - b$$ Since $|x - a| \ge 0$, for both to hold we must have $b-y \ge 0$ and $y-b \ge 0$, so $b = y$. Hence $|x - a| \le 0 \Rightarrow x = a$. Thus $(x,y,z) = (a,b,c)$, proving antisymmetry. - Transitive: Suppose $(x,y,z)𝓡(a,b,c)$ and $(a,b,c)𝓡(p,q,r)$. Then $$|x-a| \le b - y, \quad z = c$$ $$|a-p| \le q - b, \quad c = r$$ By triangle inequality: $$|x-p| \le |x-a| + |a-p| \le (b - y) + (q - b) = q - y$$ And $z = c = r$, so $$(x,y,z)𝓡(p,q,r)$$ Thus, 𝓡 is transitive. Since 𝓡 is reflexive, antisymmetric, and transitive, 𝓡 is an order relation on $\mathbb{R}^3$. 2. **Determine whether this order is total on ℝ³.** An order is total if for any distinct $(x,y,z)$ and $(a,b,c)$, either $(x,y,z)𝓡(a,b,c)$ or $(a,b,c)𝓡(x,y,z)$. Consider $(x,y,z)$ and $(a,b,c)$ with $z \neq c$. Since $𝓡$ requires $z=c$ to hold, neither relates the other. Hence, 𝓡 is not total on $\mathbb{R}^3$. **Exercise 3** 1. **Check pairs for the relation $x𝓡y \iff \frac{2x + y}{3} \in \mathbb{N}$.** - For $7𝓡5$ $$\frac{2(7) + 5}{3} = \frac{14 + 5}{3} = \frac{19}{3} = 6.333... \notin \mathbb{N}$$ So $7𝓡5$ is false. - For $6𝓡9$ $$\frac{2(6) + 9}{3} = \frac{12 + 9}{3} = \frac{21}{3} = 7 \in \mathbb{N}$$ So $6𝓡9$ is true. - For $4𝓡4$ $$\frac{2(4) + 4}{3} = \frac{8 + 4}{3} = \frac{12}{3} = 4 \in \mathbb{N}$$ So $4𝓡4$ is true. 2. **Prove 𝓡 is an equivalence relation on $\mathbb{N}$.** An equivalence relation must be reflexive, symmetric, and transitive. - Reflexive: For any $x \in \mathbb{N}$, $$\frac{2x + x}{3} = x \in \mathbb{N}$$ So $x𝓡x$ holds. - Symmetric: Suppose $x𝓡y$, meaning $$\frac{2x + y}{3} = k \in \mathbb{N}$$ We want to check if $y𝓡x$: $$\frac{2y + x}{3}$$ From $k = \frac{2x + y}{3}$, multiply both sides by 3: $$3k = 2x + y$$ Rearrange for $y$: $$y = 3k - 2x$$ Substitute into $\frac{2y + x}{3}$: $$\frac{2(3k - 2x) + x}{3} = \frac{6k - 4x + x}{3} = \frac{6k - 3x}{3} = 2k - x$$ Since $k,x \in \mathbb{N}$, $2k - x$ is an integer but may not be natural (depends on $k,x$). For symmetry to hold, $2k - x$ must be natural. Check examples: - For $x=6,y=9$ as above, $k=7$ and $2k - x=14-6=8 \in \mathbb{N}$, so $y𝓡x$ holds. - For $x=7,y=5$, $x𝓡y$ false, so no contradiction. However, this argument is not sufficient to conclude symmetry definitively for all $x,y$. Instead, rewrite the relation: $x𝓡y \iff 2x + y \equiv 0 \pmod{3}$ since membership in $\mathbb{N}$ is about divisibility by 3. Then $x𝓡y \iff 2x + y \equiv 0 \pmod{3}$. For symmetry check: $$2x + y \equiv 0 \pmod{3} \implies 2y + x \equiv ? \pmod{3}$$ Multiply the first by 2: $$2(2x + y) = 4x + 2y \equiv 0 \pmod{3}$$ Since $4x \equiv x \pmod{3}$ (because $4 \equiv 1$ mod 3), $$x + 2y \equiv 0 \pmod{3}$$ So $2y + x \equiv 0 \pmod{3}$ iff $x + 2y \equiv 0 \pmod{3}$. Hence symmetry does not hold in general. Therefore, 𝓡 is not symmetric and thus not an equivalence relation. 3. **Find the equivalence class $\bar{x}$ for an arbitrary $x \in \mathbb{N}$.** Since 𝓡 is not an equivalence relation, equivalence classes do not exist. **Exercise 4** 1. **Analyze and describe the function:** $$f(x) = \frac{x + 1}{2x - 1}, \quad x \in \mathbb{R} \setminus \left\{ \frac{1}{2} \right\}$$ - Domain: All real numbers except $x = \frac{1}{2}$ where denominator is zero. - Vertical asymptote at $x = \frac{1}{2}$. - Horizontal asymptote: $$\lim_{x \to \pm \infty} f(x) = \lim_{x \to \pm \infty} \frac{x + 1}{2x - 1} = \frac{1}{2}$$ - $f(x)$ is a rational function with a vertical asymptote. - To find intercepts: - $y$-intercept at $x=0$: $$f(0) = \frac{0 + 1}{0 - 1} = -1$$ - $x$-intercept by setting numerator zero: $$x + 1 = 0 \implies x = -1$$ - $f(-1) = 0$ (crosses x-axis). **Summary:** - Domain: $\mathbb{R} \setminus \{\frac{1}{2}\}$. - Vertical asymptote: $x = \frac{1}{2}$. - Horizontal asymptote: $y = \frac{1}{2}$. - Intercepts: $(0,-1)$ and $(-1,0)$. These complete your exercises.